Answers!
hi
the question is subject to y(0)=2
Comment
Hi William,
Can you please check the question again?
From your question,
(2xydx+dy)e^x^2=0
(2xy dx + dy) = 0
2xy dx = - dy
2x dx = -1/y dy
And then integrating both sides
x2 = - log y + C
x2 - C = - log y
- x2 + C = log y
log y = - x2 + C
y = e(-x^2 + C)
y = e-x^2 eC
y = k e-x^2
Answer the question...
hi
the question is subject to y(0)=2
Comment
Hi William,
Can you please check the question again?
From your question,
(2xydx+dy)e^x^2=0
(2xy dx + dy) = 0
2xy dx = - dy
2x dx = -1/y dy
And then integrating both sides
x2 = - log y + C
x2 - C = - log y
- x2 + C = log y
log y = - x2 + C
y = e(-x^2 + C)
y = e-x^2 eC
y = k e-x^2